How do you simplify root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) ?
2 Answers
Explanation:
Let:
x = root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))
Then:
x^3 = (135+78sqrt(3))+(135-78sqrt(3))+3(root(3)(135+78sqrt(3))root(3)(135-78sqrt(3)))(root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)))
color(white)(x^3) = 270+3(root(3)(135^2-78^2(3)))x
color(white)(x^3) = 270+3(root(3)(18225-18252))x
color(white)(x^3) = 270+3(root(3)(-27))x
color(white)(x^3) = 270-9x
So:
x^3+9x-270 = 0
By the rational roots theorem, any rational roots of this cubic are expressible in the form
So the first few possible rational zeros are:
+-1, +-2, +-3, +-5, +-6, +-9, +-10,...
By Descartes' Rule of Signs, we can tell that the cubic has exactly one positive real root and no negative real zeros, so try the possible positive rational zeros.
We find:
(color(blue)(6))^3+9(color(blue)(6))-270 = 216+54-270 = 0
So
x^3+9x-270 = (x-6)(x^2+6x+45)
The remaining quadratic has no real zeros (as we can see by looking at its discriminant).
The non-real zeros are actually:
omega root(3)(135+78sqrt(3))+omega^2 root(3)(135-78sqrt(3))
and:
omega^2 root(3)(135+78sqrt(3))+omega root(3)(135-78sqrt(3))
where
So:
root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6
Explanation:
Let us note that,
So, from this we guess :
If
By Trial and Error Method,
Therefore,
Consequently,
Finally,
as Respected George C. Sir has already derived!